Ok guys, I have a problem with proving this result... I have a random variable $Z$ that can take the values $[1, 2, 3]$ with probability $[\pi_1, \pi_2, \pi_3]$. How can I prove that $\mathbb{E}[Z]=2$ if and only if $\pi_3=\pi_1$? I tried to do the algebra, but no helpful cancelation happens...
2026-04-05 00:17:47.1775348267
Expected value of a discrete random variable
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Your probabilities are $\pi_1, 1-\pi_1-\pi_3$ and $\pi_3$ since the sum of the probabilities is $1$ Then$E(X)=\pi_1+2-2\pi_1-2\pi_3+3\pi_3=2$ and the result follows immediately